In the given circuit, what is the potential d | vedclass

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(A) The circuit is open between $X$ and $Y$.Since the circuit is open, no current flows through the circuit $(I = 0 \, A)$.Applying Kirchhoff's Voltag

Vedclass · Accepted Answer

$120$

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(A) The circuit is open between $X$ and $Y$.Since the circuit is open, no current flows through the circuit $(I = 0 \, A)$.Applying Kirchhoff's Voltage Law $(KVL)$ starting from point $X$ to point $Y$ along the path of the battery:The potential at $X$ is $V_X$.Moving through the resistor $(20 \, \Omega)$, there is no voltage drop because $I = 0$.Moving through the battery from the positive terminal to the negative terminal, the potential decreases by $120 \, V$.Moving through the second resistor $(20 \, \Omega)$, there is no voltage drop.Thus, the equation is: $V_X - 120 \, V = V_Y$.Rearranging the terms, we get $V_X - V_Y = 120 \, V$.Therefore, the potential difference between $X$ and $Y$ is $120 \, V$.

In the given circuit, what is the potential difference between $X$ and $Y$ in $V$?

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